Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 4 - Section 4.2 - Solving Systems of Linear Equations in Three Variables - Exercise Set - Page 221: 40

Answer

$(1,1,-7)$

Work Step by Step

Given Equations are $x+3y+z=-3$ Equation $(1)$ $-x+y+2z=-14$ Equation $(2)$ $3x+2y-z=12$ Equation $(3)$ Adding Equation $(1)$ and Equation $(2)$ $x+3y+z-x+y+2z=-3-14$ $4y+3z=-17$ Equation $(4)$ Adding Equation $(1)$ and Equation $(3)$ $x+3y+z+3x+2y-z=-3+12$ $4x+5y=9$ Equation $(5)$ Multiplying Equation $(2)$ by $3$ then adding with Equation $(3)$ $3(-x+y+2z)+(3x+2y-z)=3(-14)+12$ $-3x+3y+6z+3x+2y-z=-42+12$ $5y+5z=-30$ $y+z=-6$ Equation $(6)$ Multiplying Equation $(6)$ by $-3$ then adding with Equation $(4)$ $-3(y+z)+4y+3z=-3(-6)-17$ $-3y-3z+4y+3z=18-17$ $y=1$ Substituting $y$ value in Equation $(6)$ we get, $y+z=-6$ $1+z=-6$ $z=-6-1$ $z=-7$ Substituting $y$ value in Equation $(5)$ we get, $4x+5y=9$ $4x+5(1)=9$ $4x=9-5$ $4x=4$ $x=1$ Solution $(1,1,-7)$ Substituting $x,y,z$ values in $\frac{x}{2} + \frac{y}{3} + \frac{z}{9} = \frac{1}{18}$ $\frac{1}{2} + \frac{1}{3} - \frac{7}{9} = \frac{1}{18}$ $\frac{9+6-14}{18} = \frac{1}{18}$ $\frac{15-14}{18} = \frac{1}{18}$ $\frac{1}{18} = \frac{1}{18}$
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